Solve (2+1/2t)^2+4(frac{sqrt{3}}{2}t+sqrt{3})^2=4

$$( 2 + \frac { 1 } { 2 } t ) ^ { 2 } + 4 ( \frac { \sqrt { 3 } } { 2 } t + \sqrt { 3 } ) ^ { 2 } = 4$$

Solve for t

$$t=\frac{4\sqrt{10}-28}{13}\approx -1.180837643$$

$$t=\frac{-4\sqrt{10}-28}{13}\approx -3.126854665$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\frac{1}{2}t\right)^{2}.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\sqrt{3}}{2}t+\sqrt{3}\right)^{2}=4$$

Express \frac{\sqrt{3}}{2}t as a single fraction.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\sqrt{3}t}{2}+\sqrt{3}\right)^{2}=4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{\sqrt{3}t}{2}+\sqrt{3}\right)^{2}.

$$4+2t+\frac{1}{4}t^{2}+4\left(\left(\frac{\sqrt{3}t}{2}\right)^{2}+2\times \frac{\sqrt{3}t}{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

To raise \frac{\sqrt{3}t}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+2\times \frac{\sqrt{3}t}{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

Express 2\times \frac{\sqrt{3}t}{2} as a single fraction.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\frac{2\sqrt{3}t}{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

Cancel out 2 and 2.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\sqrt{3}t\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

The square of \sqrt{3} is 3.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\sqrt{3}t\sqrt{3}+3\right)=4$$

Use the distributive property to multiply 4 by \frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\sqrt{3}t\sqrt{3}+3.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+4\sqrt{3}t\sqrt{3}+12=4$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+4\times 3t+12=4$$

Expand \left(\sqrt{3}t\right)^{2}.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{\left(\sqrt{3}\right)^{2}t^{2}}{2^{2}}+4\times 3t+12=4$$

The square of \sqrt{3} is 3.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{3t^{2}}{2^{2}}+4\times 3t+12=4$$

Calculate 2 to the power of 2 and get 4.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{3t^{2}}{4}+4\times 3t+12=4$$

Express 4\times \frac{3t^{2}}{4} as a single fraction.

$$4+2t+\frac{1}{4}t^{2}+\frac{4\times 3t^{2}}{4}+4\times 3t+12=4$$

Cancel out 4 and 4.

$$4+2t+\frac{1}{4}t^{2}+3t^{2}+4\times 3t+12=4$$

Multiply 4 and 3 to get 12.

$$4+2t+\frac{1}{4}t^{2}+3t^{2}+12t+12=4$$

Combine \frac{1}{4}t^{2} and 3t^{2} to get \frac{13}{4}t^{2}.

$$4+2t+\frac{13}{4}t^{2}+12t+12=4$$

Combine 2t and 12t to get 14t.

$$4+14t+\frac{13}{4}t^{2}+12=4$$

Add 4 and 12 to get 16.

$$16+14t+\frac{13}{4}t^{2}=4$$

Subtract 4 from both sides.

$$16+14t+\frac{13}{4}t^{2}-4=0$$

Subtract 4 from 16 to get 12.

$$12+14t+\frac{13}{4}t^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{13}{4}t^{2}+14t+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{13}{4} for a, 14 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-14±\sqrt{14^{2}-4\times \frac{13}{4}\times 12}}{2\times \frac{13}{4}}$$

Square 14.

$$t=\frac{-14±\sqrt{196-4\times \frac{13}{4}\times 12}}{2\times \frac{13}{4}}$$

Multiply -4 times \frac{13}{4}.

$$t=\frac{-14±\sqrt{196-13\times 12}}{2\times \frac{13}{4}}$$

Multiply -13 times 12.

$$t=\frac{-14±\sqrt{196-156}}{2\times \frac{13}{4}}$$

Add 196 to -156.

$$t=\frac{-14±\sqrt{40}}{2\times \frac{13}{4}}$$

Take the square root of 40.

$$t=\frac{-14±2\sqrt{10}}{2\times \frac{13}{4}}$$

Multiply 2 times \frac{13}{4}.

$$t=\frac{-14±2\sqrt{10}}{\frac{13}{2}}$$

Now solve the equation t=\frac{-14±2\sqrt{10}}{\frac{13}{2}} when ± is plus. Add -14 to 2\sqrt{10}.

$$t=\frac{2\sqrt{10}-14}{\frac{13}{2}}$$

Divide -14+2\sqrt{10} by \frac{13}{2} by multiplying -14+2\sqrt{10} by the reciprocal of \frac{13}{2}.

$$t=\frac{4\sqrt{10}-28}{13}$$

Now solve the equation t=\frac{-14±2\sqrt{10}}{\frac{13}{2}} when ± is minus. Subtract 2\sqrt{10} from -14.

$$t=\frac{-2\sqrt{10}-14}{\frac{13}{2}}$$

Divide -14-2\sqrt{10} by \frac{13}{2} by multiplying -14-2\sqrt{10} by the reciprocal of \frac{13}{2}.

$$t=\frac{-4\sqrt{10}-28}{13}$$

The equation is now solved.

$$t=\frac{4\sqrt{10}-28}{13} t=\frac{-4\sqrt{10}-28}{13}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\frac{1}{2}t\right)^{2}.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\sqrt{3}}{2}t+\sqrt{3}\right)^{2}=4$$

Express \frac{\sqrt{3}}{2}t as a single fraction.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\sqrt{3}t}{2}+\sqrt{3}\right)^{2}=4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{\sqrt{3}t}{2}+\sqrt{3}\right)^{2}.

$$4+2t+\frac{1}{4}t^{2}+4\left(\left(\frac{\sqrt{3}t}{2}\right)^{2}+2\times \frac{\sqrt{3}t}{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

To raise \frac{\sqrt{3}t}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+2\times \frac{\sqrt{3}t}{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

Express 2\times \frac{\sqrt{3}t}{2} as a single fraction.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\frac{2\sqrt{3}t}{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

Cancel out 2 and 2.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\sqrt{3}t\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)=4$$

The square of \sqrt{3} is 3.

$$4+2t+\frac{1}{4}t^{2}+4\left(\frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\sqrt{3}t\sqrt{3}+3\right)=4$$

Use the distributive property to multiply 4 by \frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+\sqrt{3}t\sqrt{3}+3.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+4\sqrt{3}t\sqrt{3}+12=4$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{\left(\sqrt{3}t\right)^{2}}{2^{2}}+4\times 3t+12=4$$

Expand \left(\sqrt{3}t\right)^{2}.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{\left(\sqrt{3}\right)^{2}t^{2}}{2^{2}}+4\times 3t+12=4$$

The square of \sqrt{3} is 3.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{3t^{2}}{2^{2}}+4\times 3t+12=4$$

Calculate 2 to the power of 2 and get 4.

$$4+2t+\frac{1}{4}t^{2}+4\times \frac{3t^{2}}{4}+4\times 3t+12=4$$

Express 4\times \frac{3t^{2}}{4} as a single fraction.

$$4+2t+\frac{1}{4}t^{2}+\frac{4\times 3t^{2}}{4}+4\times 3t+12=4$$

Cancel out 4 and 4.

$$4+2t+\frac{1}{4}t^{2}+3t^{2}+4\times 3t+12=4$$

Multiply 4 and 3 to get 12.

$$4+2t+\frac{1}{4}t^{2}+3t^{2}+12t+12=4$$

Combine \frac{1}{4}t^{2} and 3t^{2} to get \frac{13}{4}t^{2}.

$$4+2t+\frac{13}{4}t^{2}+12t+12=4$$

Combine 2t and 12t to get 14t.

$$4+14t+\frac{13}{4}t^{2}+12=4$$

Add 4 and 12 to get 16.

$$16+14t+\frac{13}{4}t^{2}=4$$

Subtract 16 from both sides.

$$14t+\frac{13}{4}t^{2}=4-16$$

Subtract 16 from 4 to get -12.

$$14t+\frac{13}{4}t^{2}=-12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{13}{4}t^{2}+14t=-12$$

Divide both sides of the equation by \frac{13}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{13}{4}t^{2}+14t}{\frac{13}{4}}=-\frac{12}{\frac{13}{4}}$$

Dividing by \frac{13}{4} undoes the multiplication by \frac{13}{4}.

$$t^{2}+\frac{14}{\frac{13}{4}}t=-\frac{12}{\frac{13}{4}}$$

Divide 14 by \frac{13}{4} by multiplying 14 by the reciprocal of \frac{13}{4}.

$$t^{2}+\frac{56}{13}t=-\frac{12}{\frac{13}{4}}$$

Divide -12 by \frac{13}{4} by multiplying -12 by the reciprocal of \frac{13}{4}.

$$t^{2}+\frac{56}{13}t=-\frac{48}{13}$$

Divide \frac{56}{13}, the coefficient of the x term, by 2 to get \frac{28}{13}. Then add the square of \frac{28}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{56}{13}t+\left(\frac{28}{13}\right)^{2}=-\frac{48}{13}+\left(\frac{28}{13}\right)^{2}$$

Square \frac{28}{13} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{56}{13}t+\frac{784}{169}=-\frac{48}{13}+\frac{784}{169}$$

Add -\frac{48}{13} to \frac{784}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{56}{13}t+\frac{784}{169}=\frac{160}{169}$$

Factor t^{2}+\frac{56}{13}t+\frac{784}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{28}{13}\right)^{2}=\frac{160}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{28}{13}\right)^{2}}=\sqrt{\frac{160}{169}}$$

Simplify.

$$t+\frac{28}{13}=\frac{4\sqrt{10}}{13} t+\frac{28}{13}=-\frac{4\sqrt{10}}{13}$$

Subtract \frac{28}{13} from both sides of the equation.

$$t=\frac{4\sqrt{10}-28}{13} t=\frac{-4\sqrt{10}-28}{13}$$